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3 years ago

If a randomly selected customer was satisfied with his/her meal, what is the probability that the customer ordered fish

Mathematics

1 answer:

vodka [1.7K]3 years ago

7 0

Answer:

87.5%

Step-by-step explanation:

Complete question -

75% of the people who eat at Doug's Diner order fish and 90% of the people who order fish also order fries. Given that 80% of all people who eat at Doug's order fries, what is the probability that a randomly chosen customer orders fries without fish?

Solution

Given

People who order fish at the Doug's Diner = 75%

Out of this 75%, 90% also order fries which means that nearly 67.5% ordered fries with the fish.

Given that, 80% of people dining at Doug's Diner order fries

Then percentage or people who ordered only fires and no fish = 80% - 67.5% = 12.5%

The probability of customers who ordered fish = 100 - 12.5% = 87.5%

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Answer:

Step-by-step explanation:

<u>Statements </u> <u> Reasons</u>

1) QS =42 Given

2) QR + RS = QS Segment Addition Postulate

3) (x + 3) + 2x = 42 Substitution Property

4) 3x + 3 = 42 Simplify

5) 3x = 39 Subtraction Property of Equality

6) x=13 Division Property of Equality

Explanation:

We have given QS=42. We have to prove that x=13

We will use Segment Addition Postulate which states that given 2 points Q and S, a third point R lies on the line segment QS if and only if the distances between the points satisfy the equation QR + RS = QS.

Then we will substitute the values in the defined postulate.

where QR= x+3

RS=2x

QS=42

QR+RS=QS

(x+3)+2x= 42

Now simplify the expression by opening the brackets.

x+3+2x=42

3x+3=42

Now subtract 3 from both sides.

3x+3-3=42-3

3x=39

divide both sides by 3.

3x/3 =39/3

x=13..

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2 years ago

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This is further explained below.

<h3>What are gridlines?</h3>

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